How to solve this type of exercises $\sqrt{x^6+x^5-2x^3+O(x^2)}$

Question

I have a simulation test with this type of exercise, asymptotic expansion:

$$\sqrt{x^6+x^5-2x^3+O(x^2)}$$

with

$$ x\rightarrow \infty$$

I have studied the theory of Landau's symbols but I have no idea about how to solve.

Can someone please explain me how to do that?

No. By definition the square root of a value squared gives the value. — user21820, Jun 22 '15 at 12:02
@ChristianGiupponi: You are mixing up addition and multiplication. $a\sqrt{b}=\sqrt{a^2b}$, whereas what you tried to do is $a\sqrt{b}=\sqrt{a^2+b}$. — Meni Rosenfeld, Jun 22 '15 at 12:18
Ok, thanks, now is much more clear :) Can you give me another hint to understand how to solve? Just the first comment isn't enough for me in this case — Christian Giupponi, Jun 22 '15 at 12:22

score 0 · Answer 1 · edited Jun 22 '15 at 13:01

0

Write $$\sqrt{x^6+x^5-2x^3+O(x^2)}=$$ $$=\sqrt{x^6+x^5-2x^3}+\sqrt{x^6+x^5-2x^3+O(x^2)}-\sqrt{x^6+x^5-2x^3}=$$ $$=\sqrt{x^6+x^5-2x^3}+\frac{O(x^2)}{\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6+x^5-2x^3}}.$$ But $\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6+x^5-2x^3}\sim 2x^3$, hence $$\frac{O(x^2)}{\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6+x^5-2x^3}}=O(x^{-1})$$ thus $$\sqrt{x^6+x^5-2x^3+O(x^2)}=\sqrt{x^6+x^5-2x^3}+O(x^{-1})$$

edited Jun 22 '15 at 13:01

Meni Rosenfeld

4,013

answered Jun 22 '15 at 12:22

Fabio Lucchini

16,054

why did you do $\frac{O(x^2)}{\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6+x^5-2x^3}}$ – Christian Giupponi Jun 25 '15 at 08:59
@ChristianGiupponi: He did it because it brought him closer to the final answer - he replaced subtraction between two similar terms, which by itself is difficult to evaluate, with a division between two terms with clear growth. To see that it's correct, try multiplying the denominator with the expression he's replacing. – Meni Rosenfeld Jun 25 '15 at 17:51

How to solve this type of exercises $\sqrt{x^6+x^5-2x^3+O(x^2)}$

1 Answers1